Didier wrote:
>> Square root of 2 is about 1,414 or about 3,01 dB.
>
> I am always confused when considering noise, is it 10*log(p1/p0) or
> 20*log(p1/p0)?
A moment's reflection on why the 10 log vs. 20 log, might help.
The conversion from a power ratio to dB is:
dB = 10 log (P1/P2)
Remember that Power = VxV/R, so:
P1/P2 = (V1xV1)/(V2xV2), the R's cancelling.
So,
dB = 10 log [(V1^2)/V2^2)] or, 10 log[(V1/V2)^2]
If we want to express this as a ratio of voltages, rather than a
ratio of powers (there's a pun in there somewhere ;-),
we need to take the square root of (V1/V2)^2 outside of the log.
To do this, we need to remember that log[X^2] = 2 log X, so:
dB = 10 log[(V1/V2)^2] = 20 log[V1/V2]
A couple of things to note:
1) dB's are dB's. 3dB represents the doubling of a power ratio,
6dB represents the doubling of a voltage ratio.
2) Convention says that if -dB's are loss, and +dB's are gain, but that
is just convention.
-Chuck Harris
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